Trigonometric Equations Question 172
Question: A tower of height b subtends an angle at a point O on the level of the foot of the tower and at a distance a from the foot of the tower. If a pole mounted on the tower also subtends an equal angle at O, the height of the pole is
[MP PET 1993, 2004]
Options:
A) $ b,( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} ) $
B) $ b,( \frac{a^{2}+b^{2}}{a^{2}-b^{2}} ) $
C) $ a,( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} ) $
D) $ a,( \frac{a^{2}+b^{2}}{a^{2}-b^{2}} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \tan \alpha =\frac{b}{a} $ , $ \tan 2\alpha =\frac{2(b/a)}{1-{{(b/a)}^{2}}}=\frac{p+b}{a} $
$ \Rightarrow $ $ \frac{2ba}{a^{2}-b^{2}}=\frac{p+b}{a}\Rightarrow \frac{2ba^{2}-a^{2}b+b^{3}}{a^{2}-b^{2}}=p $
$ \Rightarrow $ $ p=\frac{b(a^{2}+b^{2})}{(a^{2}-b^{2})} $ .
BETA
